Constructing Poncelet Quadrilaterals using inversive geometry

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In the diagram above, you are given a circle with centre at \(O\) and a point \(R\) on the circle. Inside the circle is a point \(P\). Given a pair of perpendicular lines passing through \(P\), let \(A\), \(B\), \(C\) and \(D\) be the points where these lines intersect the circle.

We showed in Question 1 of Homework 2 that the midpoint \(M_1\) of \(AB\) moves along a fixed circle as the pair of perpendicular lines through \(P\) is rotated. You can see this in the picture by moving the point \(D\) around the blue circle. The point \(M_1\) will move around the green circle, and so will the other midpoints \(M_2\), \(M_3\) and \(M_4\) for the same reason.

Now draw tangents to the blue circle at points \(A\), \(B\), \(C\) and \(D\), and let \(XYZW\) be the quadrilateral formed by the intersection of these tangent lines. In Question 6 on Tutorial 9, you were asked to show that the vertices \(X\), \(Y\), \(Z\) and \(W\) lie on a fixed circle. As you can see from the diagram, \(X\) is the inverse of \(M_1\) with respect to the blue circle, by the tangent construction of the inverse. Therefore \(X\) lies on the inverse of the green circle, and the same is true for \(Y\), \(Z\) and \(W\).

This gives a method for constructing Poncelet Quadrilaterals. By moving the points \(O\), \(P\), \(R\) and \(D\) in the diagram, you can see explicitly that we always have a closed quadrilateral which is tangent to the blue circle and which has vertices on the red circle.